Chapter 7 The Mathematics of Maybe: Introduction to Probability (Class 9 - Latest Maths NCERT (Ganita Manjari I) Solutions)

(i) Event: The next Monday will come after Sunday.

Rank: $1$

Label: Certain

Reason: The sequence of the days of the week is a fixed cycle. According to the standard calendar used in India, Monday always follows Sunday. There is no possibility of any other day occurring between them.

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(ii) Event: It will snow in Mumbai in July.

Rank: $0$

Label: Impossible

Reason: Mumbai is a coastal city in India with a tropical climate. During the month of July, the city experiences the peak of the southwest monsoon with high temperatures and heavy rainfall. The atmospheric conditions are never cold enough for snowfall to occur.

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(iii) Event: An elephant will walk through your classroom today.

Rank: $0$

Label: Impossible

Reason: Classrooms are strictly controlled indoor environments located within school buildings, typically in urban or residential areas. It is physically and logically impossible for a large wild animal like an elephant to enter a modern school infrastructure and walk through a specific classroom during a normal school day.

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(iv) Event: You will greet at least one friend at school tomorrow.

Rank: $0.9$

Label: More likely

Reason: Since you are a student and have a regular social circle at school, it is highly probable that you will meet and greet your friends. While it is not "Certain" (as you or your friends could be absent), the frequency of this occurrence makes it very more likely.

Given:

Total number of sweets in the sample ($n$) $= 30$

Number of red sweets $= 10$

Number of green sweets $= 8$

Number of yellow sweets $= 7$

Number of blue sweets $= 5$

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To Find:

(i) Probability of picking a green sweet.

(ii) Estimated number of yellow sweets in a bag of $600$.

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Solution:

(i) Probability of green sweet:

Simplifying the fraction:

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(ii) Estimation of yellow sweets:

First, find the probability of a yellow sweet from the sample:

To estimate the number of yellow sweets in the whole bag of $600$:

Therefore, there are likely to be $140$ yellow sweets in the large bag.

Given:

Sample size ($n$) $= 40$

Students preferring Arts Club $= 11$

Students preferring Sports Club $= 9$

Total students in school $= 800$

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Solution:

(i) Probability for Arts Club:

As a decimal, $P(\text{Arts}) = 0.275$.

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(ii) Estimation for Sports Club:

Probability of choosing a student who prefers Sports from the sample:

Estimated number of students in the whole school ($800$):

Therefore, approximately $180$ students in the school are likely to prefer the Sports Club.

Note: Since this is an activity-based question, we will assume a sample set of results for the purpose of the calculation.

Assumed Results: Let's say out of $20$ tosses, we obtained Heads $12$ times and Tails $8$ times.

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Solution:

(i) Number of Heads:

Based on our assumed data, the number of heads obtained is $12$.

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(ii) Number of Tails:

Based on our assumed data, the number of tails obtained is $8$.

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(iii) Experimental Probability of Heads:

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(iv) Probability of getting tails on the next toss:

In Experimental Probability, we use the results of past trials to predict the next one.

Note: Theoretically, for a fair coin, the probability of getting tails on any single toss is always $0.5$, but based on this experiment, the probability is $0.4$.

Note: Since this is an activity-based experiment, the results will vary for everyone. For this solution, we will assume a sample set of observations from $100$ tosses.

Assumed Observations:

Total number of trials (tosses) $= 100$

1. Number of times cup landed on its Bottom $= 8$

2. Number of times cup landed on its Top (Upside down) $= 12$

3. Number of times cup landed on its Side $= 80$

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Solution:

The experimental probability of an event is given by:

$P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}}$

1. Probability of landing on the Bottom:

2. Probability of landing on the Top:

3. Probability of landing on the Side:

To Find:

The theoretical probability of getting an even number on a fair $6$-sided die.

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Solution:

When a fair $6$-sided die is rolled, the sample space (total possible outcomes) is:

$S = \{1, 2, 3, 4, 5, 6\}$

Total number of outcomes $n(S) = 6$

Let $E$ be the event of getting an even number. The favorable outcomes are:

$E = \{2, 4, 6\}$

Number of favorable outcomes $n(E) = 3$

The probability $P(E)$ is given by:

Therefore, the probability of getting an even number is $\frac{1}{2}$.

Given:

Total number of rolls $= 12$

Number of times '$3$' occurred $= 3$

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Solution:

(i) Experimental Probability:

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(ii) Theoretical Probability:

For a fair $6$-sided die, there is only one '$3$' out of $6$ total faces.

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(iii) Reasoning and Expectation:

Reason for difference: Experimental probability is based on a limited number of trials and is subject to random chance and fluctuations. In a small sample size like $12$ rolls, the outcomes may not reflect the mathematical ideal (theoretical probability).

Expectation for larger trials: According to the Law of Large Numbers, as the number of trials increases ($60, 600,$ or $6000$), the experimental probability tends to get closer and closer to the theoretical probability.

For example, if we roll the die $6000$ times, we would expect the outcome '$3$' to occur approximately $\frac{1}{6}$ of the time, which is about $1000$ times, making the experimental probability very close to $0.167$.

Solution:

When a single 6-sided die is rolled, the numbers that can appear on the top face are $1, 2, 3, 4, 5$ or $6$.

The sample space ($S$) is the set of all possible outcomes.

The total number of elements in the sample space is denoted by $n(S)$.

Therefore, the total number of possible outcomes in the sample space is 6.

Definition:

A sample space (denoted by $S$) is the set of all possible outcomes of a random experiment. Each element in the set is called a sample point.

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(i) Rolling a die and tossing a coin together:

When a $6$-sided die is rolled, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. When a coin is tossed, the possible outcomes are $\{H, T\}$, where $H$ stands for Heads and $T$ for Tails.

The sample space $S$ consists of all possible ordered pairs $(\text{die outcome, coin outcome})$ as shown in the table below:

Die Outcome	Coin Outcome	Sample Point (Die, Coin)
1	H, T	(1, H), (1, T)
2	H, T	(2, H), (2, T)
3	H, T	(3, H), (3, T)
4	H, T	(4, H), (4, T)
5	H, T	(5, H), (5, T)
6	H, T	(6, H), (6, T)

Thus, the sample space is:

$S = \{(1, H), (1, T), (2, H), (2, T), (3, H), (3, T), (4, H), (4, T), (5, H), $$ (5, T), (6, H), (6, T)\}$

Total number of outcomes $n(S) = 6 \times 2 = 12$.

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(ii) Choosing a random integer between $-5$ and $+5$:

In mathematics, the term between typically refers to the set of integers strictly greater than $-5$ and strictly less than $5$ (excluding the endpoints).

The integers lying in this range are:

$S = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$

Total number of outcomes $n(S) = 9$.

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(iii) A box containing $5$ green and $7$ red balls. One ball is drawn at random:

To ensure that each outcome is equally likely, we treat each ball as a distinct object. Let the $5$ green balls be $G_1, G_2, G_3, G_4, G_5$ and the $7$ red balls be $R_1, R_2, R_3, R_4, R_5, R_6, R_7$.

The sample space $S$ consists of all individual balls that can be drawn:

$S = \{G_1, G_2, G_3, G_4, G_5, R_1, R_2, R_3, R_4, R_5, R_6, R_7\}$

Total number of outcomes $n(S) = 5 + 7 = 12$.

Note: If the sample space is required only in terms of the observed color of the ball, it would be $S = \{\text{Green, Red}\}$. However, for probability calculations, listing all $12$ distinct outcomes is the standard mathematical approach.

Solution:

(i) Sample space $S$:

The sample space contains all possible combinations of one snack and one drink.

$S = \{(\text{Samosa, Chai}), (\text{Samosa, Lassi}), (\text{Pakora, Chai}), $$ (\text{Pakora, Lassi}), (\text{Bhaji, Chai}), (\text{Bhaji, Lassi})\}$

The total number of possible combinations is 6.

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(ii) Event ‘Selecting Samosa as a snack’:

Let $E$ be the event of selecting a Samosa. This event consists of all outcomes where the snack is Samosa regardless of the drink chosen.

$E = \{(\text{Samosa, Chai}), (\text{Samosa, Lassi})\}$

Given:

Basket $A$ contains: $1$ Apple ($A$), $2$ Oranges ($O_1$ and $O_2$). Total fruits in $A = 3$.

Basket $B$ contains: $1$ Banana ($Ba$), $1$ Mango ($M$). Total fruits in $B = 2$.

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(i) Tree Diagram:

The tree diagram starts with three branches for the first pick (Basket $A$) and each of those branches splits into two for the second pick (Basket $B$).

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(ii) Sample Space:

The sample space $S$ is the set of all possible ordered pairs $(\text{Fruit from } A, \text{Fruit from } B)$.

Total number of outcomes in the sample space is $n(S)$:

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(iii) Probability of picking one apple and one banana:

Let $E$ be the event of picking one apple and one banana.

Looking at the sample space, the only favourable outcome is $(A, Ba)$.

The probability $P(E)$ is calculated as:

Therefore, the probability of picking one apple and one banana is $\frac{1}{6}$.

Given:

Number of Red pens ($R$) $= 3$

Number of Black pens ($Bl$) $= 4$

Number of Green pens ($G$) $= 2$

Total number of pens $= 3 + 4 + 2 = 9$

The experiment is "with replacement," meaning the total count remains $9$ for both picks.

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(i) Possible outcomes and Tree Diagram:

The possible outcomes for the colours are Red ($R$), Black ($Bl$), and Green ($G$). Since two people pick, the possible outcomes for the pair of colours are: $(R, R), (R, Bl), (R, G), (Bl, R), (Bl, Bl), (Bl, G), (G, R), (G, Bl),$ and $(G, G)$.

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(ii) Probability that both pick pens of the same colour:

Let the event of both picking the same colour be $E$. The favourable outcomes are $(R, R), (Bl, Bl),$ and $(G, G)$.

Total number of outcomes in the sample space $n(S) = 9 \times 9 = 81$.

1. Probability of picking Red twice:

2. Probability of picking Black twice:

3. Probability of picking Green twice:

The total probability of picking the same colour is the sum of these probabilities:

Therefore, the probability that both pick pens of the same colour is $\frac{29}{81}$.

(i) The probability of an impossible event is $0$.

(ii) The set of all possible outcomes of a random experiment is called the sample space.

(iii) The probability of an event that is certain to happen is $1$.

(iv) Tossing a fair coin has a probability of $\frac{1}{2}$ (or $0.5$) for getting heads.

Solution:

In this survey, the number of students who like football represents the actual count of occurrences.

The relative frequency is the ratio of the frequency to the total number of students in the survey.

Therefore, the relative frequency is $0.3$ (or $\frac{3}{10}$).

Explanation:

Outcomes are said to be equally likely if each has the same chance of occurring.

(i) A driver attempts to start a car: Not equally likely. The probability depends on several factors like the fuel level, battery health, and mechanical condition. It is much more likely to start if the car is in good condition.

(ii) Tossing a fair coin once: Equally likely. A fair coin has a $50\%$ chance of landing on Heads and a $50\%$ chance of landing on Tails.

(iii) Rolling a fair 6-sided die: Equally likely. Each of the six faces $\{1, 2, 3, 4, 5, 6\}$ has an equal probability of $\frac{1}{6}$ of appearing.

(iv) Choosing a marble randomly: Not equally likely. Since the number of blue marbles ($7$) is greater than the red marbles ($3$), picking a blue marble is more likely than picking a red one.

(v) A baby is born: Equally likely. Biologically, under normal circumstances, the probability of a newborn being a boy or a girl is approximately equal (roughly $1/2$ each).

(i) Two coins tossed:

Sample Space $S = \{HH, HT, TH, TT\}$. So, $n(S) = 4$.

Event of "at least one head" $E = \{HH, HT, TH\}$. So, $n(E) = 3$.

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(ii) Card numbered 1 to 10:

Sample Space $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. So, $n(S) = 10$.

Event of "even number" $E = \{2, 4, 6, 8, 10\}$. So, $n(E) = 5$.

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(iii) Die rolled once:

Sample Space $S = \{1, 2, 3, 4, 5, 6\}$. So, $n(S) = 6$.

Event of "number greater than 4" $E = \{5, 6\}$. So, $n(E) = 2$.

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(iv) Bag of balls:

Total balls $n(S) = 3 \text{ Red} + 2 \text{ Blue} + 1 \text{ Green} = 6$.

Event "not red" means Blue or Green balls. Favourable outcomes $n(E) = 2 + 1 = 3$.

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(v) Three coins tossed:

Sample Space $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$. So, $n(S) = 8$.

Event of "exactly two heads" $E = \{HHT, HTH, THH\}$. So, $n(E) = 3$.

Given:

Total number of candies in the bag $= 3$ (Strawberry, Lemon, Mint).

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Solution:

The sample space $S$ is the set of all possible candies that can be picked.

Let $E$ be the event of picking a strawberry candy.

The probability $P(E)$ is given by the formula:

Therefore, the probability of picking a strawberry candy is $\frac{1}{3}$.

Solution:

The child has a choice between $2$ shirts ($S_1, S_2$) and $3$ pants ($P_1, P_2, P_3$). The total number of outfit combinations is $2 \times 3 = 6$.

The following table lists all possible combinations:

Shirt Colour	Pants Type	Combination (Outfit)
Red	Jeans	(Red Shirt, Jeans)
Red	Khakis	(Red Shirt, Khakis)
Red	Shorts	(Red Shirt, Shorts)
Blue	Jeans	(Blue Shirt, Jeans)
Blue	Khakis	(Blue Shirt, Khakis)
Blue	Shorts	(Blue Shirt, Shorts)

The sample space for the outfits is:
$S = \{(\text{Red, Jeans}), (\text{Red, Khakis}), (\text{Red, Shorts}), (\text{Blue, Jeans}), $$ (\text{Blue, Khakis}), (\text{Blue, Shorts})\}$.

Given:

Total number of cases (trials) $n = 1000$.

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Solution:

(i) Probability that a tyre lasts less than $4000\text{ km}$:

Number of favourable cases $= 20$.

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(ii) Probability that a tyre lasts between $4000$ and $14000\text{ km}$:

This includes the categories "$4001$ to $9000$" and "$9001$ to $14000$".

Number of favourable cases $= 210 + 325 = 535$.

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(iii) Probability that a tyre lasts more than $14000\text{ km}$:

Number of favourable cases $= 445$.

Given:

The word is PEACE. The letters are: P, E, A, C, E.

Total number of letters (cards) $n(S) = 5$.

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Solution:

(i) Probability of drawing $P$, $E$, or $C$:

Letters that are $P, E,$ or $C$ are: P (1 card), E (2 cards), and C (1 card).

Number of favourable cards $n(E_1) = 1 + 2 + 1 = 4$.

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(ii) Probability that it is not an $E$:

Total cards $= 5$. Cards that are '$E$' $= 2$.

Cards that are not '$E$' (P, A, C) $= 5 - 2 = 3$.

Therefore, the probabilities are $0.8$ and $0.6$ respectively.

Given:

The spinner is divided into $8$ equal parts numbered $1$ to $8$.

Sample Space $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$

Total number of outcomes $n(S) = 8$

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Solution:

(i) Probability that it points at $8$:

Favourable outcome $E = \{8\}$, so $n(E) = 1$.

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(ii) Probability that it points at an odd number:

Odd numbers in the set are $\{1, 3, 5, 7\}$, so $n(E) = 4$.

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(iii) Probability that it points at a number greater than $2$:

Numbers greater than $2$ are $\{3, 4, 5, 6, 7, 8\}$, so $n(E) = 6$.

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(iv) Probability that it points at a number less than $9$:

All numbers in the set are less than $9$, so $n(E) = 8$.

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(v) Probability that it points at a multiple of $3$:

Multiples of $3$ in the set are $\{3, 6\}$, so $n(E) = 2$.

Given:

Red balls ($R$) $= 4$, Blue balls ($B$) $= 5$. Total balls $= 9$.

The experiment is "without replacement" (the first ball is laid aside).

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Construction Required:

The tree diagram will have two stages (1st draw and 2nd draw).

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Solution:

(i) Probability of Red then Blue (RB):

1st draw is Red: $P(R) = \frac{4}{9}$.

Remaining balls: $3$ Red, $5$ Blue (Total $8$).

2nd draw is Blue: $P(B|R) = \frac{5}{8}$.

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(ii) Probability of 2 blue balls (BB):

1st draw is Blue: $P(B) = \frac{5}{9}$.

Remaining balls: $4$ Red, $4$ Blue (Total $8$).

2nd draw is Blue: $P(B|B) = \frac{4}{8} = \frac{1}{2}$.

Solution:

When a pair of dice is thrown, the maximum possible sum is $6 + 6 = 12$ and the minimum sum is $1 + 1 = 2$.

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Event with Probability $0$ (Impossible Event):

An event that cannot happen has a probability of $0$.

Example: "Getting a sum of $13$."

Since the highest possible sum is $12$, obtaining a sum of $13$ is impossible. Thus, $P(\text{Sum } 13) = 0$.

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Event with Probability $1$ (Certain Event):

An event that must happen has a probability of $1$.

Example: "Getting a sum that is less than $15$."

Every possible outcome ($2$ to $12$) is less than $15$. Therefore, this event is certain to happen. Thus, $P(\text{Sum } < 15) = 1$.

(i) Two dice are rolled:

Given: Two $6$-sided dice are rolled simultaneously. The desired sum is a prime number greater than $5$ (i.e., $7$ or $11$).

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Sample Space ($S$):

When two dice are rolled, the total number of outcomes is $6 \times 6 = 36$.

Die 1 \ Die 2	1	2	3	4	5	6
1	(1,1)	(1,2)	(1,3)	(1,4)	(1,5)	(1,6)
2	(2,1)	(2,2)	(2,3)	(2,4)	(2,5)	(2,6)
3	(3,1)	(3,2)	(3,3)	(3,4)	(3,5)	(3,6)
4	(4,1)	(4,2)	(4,3)	(4,4)	(4,5)	(4,6)
5	(5,1)	(5,2)	(5,3)	(5,4)	(5,5)	(5,6)
6	(6,1)	(6,2)	(6,3)	(6,4)	(6,5)	(6,6)

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Solution:

Total outcomes $n(S) = 36$.

Prime numbers greater than $5$ that can be sums are $\{7, 11\}$.

Favourable outcomes for sum $7$ are: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$.

Favourable outcomes for sum $11$ are: $(5,6), (6,5)$.

Total favourable outcomes $n(E) = 6 + 2 = 8$.

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(ii) Bag of balls (Without replacement):

Given: $4$ Red ($R_1, R_2, R_3, R_4$), $3$ Green ($G_1, G_2, G_3$), $2$ Blue ($B_1, B_2$). Total balls $= 9$. Two balls are drawn without replacement.

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Sample Space ($S$):

The total number of pairs of balls drawn without replacement is $9 \times 8 = 72$.

Possible Pairs	Count
Same Colour Red (e.g., $R_1R_2$)	$4 \times 3 = 12$
Same Colour Green (e.g., $G_1G_2$)	$3 \times 2 = 6$
Same Colour Blue (e.g., $B_1B_2$)	$2 \times 1 = 2$
Different Colours (RG, RB, GR, GB, BR, BG)	52
Total	72

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Solution:

Total outcomes $n(S) = 72$.

Number of ways to draw balls of the same colour $= 12 + 6 + 2 = 20$.

Number of ways to draw balls of different colours $n(E) = 72 - 20 = 52$.

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(iii) Three coins are tossed:

Given: $3$ coins are tossed. Find the probability that the first coin is Heads ($H$) and there are exactly two heads in total.

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Sample Space ($S$):

Outcome No.	Outcome	Exactly 2 Heads?	1st Coin Head?
1	HHH	No	Yes
2	HHT	Yes	Yes
3	HTH	Yes	Yes
4	HTT	No	Yes
5	THH	Yes	No
6	THT	No	No
7	TTH	No	No
8	TTT	No	No

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Solution:

Total outcomes $n(S) = 8$.

Favourable outcomes where 1st is H and total H count is $2$: $E = \{HHT, HTH\}$.

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(iv) Four-digit number (No repetition):

Given: Form a 4-digit number using $\{1, 2, 3, 4\}$ without repetition. Desired event: the number is even.

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Sample Space ($S$):

Total numbers formed $n(S) = 4! = 24$.

Starting with 1	Starting with 2	Starting with 3	Starting with 4
1234, 1243, 1324, 1342, 1423, 1432	2134, 2143, 2314, 2341, 2413, 2431	3124, 3142, 3214, 3241, 3412, 3421	4123, 4132, 4213, 4231, 4312, 4321

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Solution:

Even numbers must end with $2$ or $4$.

Numbers ending in $2$: $\{1342, 1432, 3142, 3412, 4132, 4312\}$ (Total $6$)

Numbers ending in $4$: $\{1234, 1324, 2134, 2314, 3124, 3214\}$ (Total $6$)

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(v) Multiple-choice test (Guessing):

Given: $3$ questions, each with options $\{A, B, C, D\}$. Exactly one correct answer per question. Desired event: exactly $2$ correct answers.

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Sample Space ($S$):

Let $C$ be a correct guess and $W$ be a wrong guess. Each question has $1$ correct and $3$ wrong possibilities. Total points in sample space $= 4 \times 4 \times 4 = 64$.

Combination Type	Example Paths	Ways to occur
3 Correct	(C, C, C)	$1 \times 1 \times 1 = 1$
2 Correct, 1 Wrong	(C,C,W), (C,W,C), (W,C,C)	$3 \times (1 \times 1 \times 3) = 9$
1 Correct, 2 Wrong	(C,W,W), (W,C,W), (W,W,C)	$3 \times (1 \times 3 \times 3) = 27$
0 Correct	(W, W, W)	$3 \times 3 \times 3 = 27$
Total		64

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Solution:

Total possible outcomes $n(S) = 64$.

Favourable outcomes (Exactly $2$ Correct):

1. Q1, Q2 correct; Q3 wrong: $1 \times 1 \times 3 = 3$ ways.

2. Q1, Q3 correct; Q2 wrong: $1 \times 3 \times 1 = 3$ ways.

3. Q2, Q3 correct; Q1 wrong: $3 \times 1 \times 1 = 3$ ways.

(i) Experiment with Replacement:

Since the ball is returned, the outcomes for the second draw are the same as the first draw $\{1, 2, 3, 4\}$.

The sample space $S_1$ can be listed in a table:

1st \ 2nd Draw	1	2	3	4
1	(1,1)	(1,2)	(1,3)	(1,4)
2	(2,1)	(2,2)	(2,3)	(2,4)
3	(3,1)	(3,2)	(3,3)	(3,4)
4	(4,1)	(4,2)	(4,3)	(4,4)

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(ii) Experiment without Replacement:

Since the first ball is not returned, the second draw cannot be the same as the first.

The sample space $S_2$ can be listed in a table:

1st \ 2nd Draw	1	2	3	4
1	—	(1,2)	(1,3)	(1,4)
2	(2,1)	—	(2,3)	(2,4)
3	(3,1)	(3,2)	—	(3,4)
4	(4,1)	(4,2)	(4,3)	—

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(iii) Sizes of Sample Spaces:

1. For the "With Replacement" experiment, there are $4$ choices for the first draw and $4$ choices for the second draw. Total outcomes $n(S_1) = 4 \times 4 = 16$.

2. For the "Without Replacement" experiment, there are $4$ choices for the first draw and $3$ choices for the second draw. Total outcomes $n(S_2) = 4 \times 3 = 12$.

Solution:

A coin has $2$ outcomes: Heads ($H$) and Tails ($T$). A set of cards numbered $1$ to $6$ has $6$ outcomes: $\{1, 2, 3, 4, 5, 6\}$.

The sample space $S$ consists of all possible pairs of coin outcomes and card numbers:

Coin \ Card	1	2	3	4	5	6
Heads (H)	(H,1)	(H,2)	(H,3)	(H,4)	(H,5)	(H,6)
Tails (T)	(T,1)	(T,2)	(T,3)	(T,4)	(T,5)	(T,6)

The elements of the sample space are:

$S = \{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), $$ (T, 4), (T, 5), (T, 6)\}$

Total number of outcomes $n(S) = 2 \times 6 = 12$.

Solution:

When three coins are tossed, the possible outcomes for the number of heads are:

1. No heads (TTT) $\rightarrow$ 0

2. One head (HTT, THT, TTH) $\rightarrow$ 1

3. Two heads (HHT, HTH, THH) $\rightarrow$ 2

4. Three heads (HHH) $\rightarrow$ 3

Thus, the valid sample space for the number of heads is $\{0, 1, 2, 3\}$. Therefore, list (iv) is the correct sample space.

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Reasons why other lists fail:

(i) $\{1, 2, 3\}$: It fails because it omits the outcome $0$ (no heads), which is a possible result when all three coins show tails.

(ii) $\{0, 1, 2\}$: It fails because it omits the outcome $3$ (three heads), which is a possible result when all three coins show heads.

(iii) $\{0, 1, 2, 3, 4\}$: It fails because it includes $4$, which is an impossible outcome. One cannot get $4$ heads from tossing only $3$ coins.

Given:

Dimensions of the rectangle: Length $= 3\text{ m}$, Breadth $= 2\text{ m}$.

Diameter of the circle inside the rectangle $= 1\text{ m}$.

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Solution:

This is a problem of geometric probability. The probability is calculated as the ratio of the area of the favourable region to the total area.

Step 1: Calculate the total area (Rectangle).

Step 2: Calculate the favourable area (Circle).

Given diameter $d = 1\text{ m}$, the radius $r = \frac{1}{2}\text{ m}$.

Step 3: Calculate the probability.

Using $\pi \approx \frac{22}{7}$:

Therefore, the probability that the dye lands inside the circle is $\frac{\pi}{24}$ or approximately $0.131$.